Forum:Math operation on collosi
I’m working on a sci-fi story, and its setting. Our observable universe is contained in a space called a domain (crude drawing here). Which itself is a part of a cosmic bubble of similar sized domains. OK, I like large numbers, so I pulled some out of thin air for the story. “Our” domain is one of 4 → 7 → 3 → 3 domains with fundamental constants that allow matter and life. But the cosmic bubble has a grand total of 2 → 3 → 2 → 2 → 2 → 2 → 2 domains; a number that makes the 4 → 7 → 3 → 3 unnoticeable. And 4 → 7 → 3 → 3 dwarfs Graham’s number. So a vast majority of the domains in the bubble are empty expanses of space. Finally, my question. Some advanced aliens are aware of the cosmology of the domains and bubbles. They reside in the nearest domain conducive to life, a space-time about a 7,000 digit number of light years away. So the human protag asks them: how far away is the next life-bearing domain? Since units don’t matter for numbers that large, I figure the genius aliens will say 2 → 3 → 2 → 2 → 2 → 2 → 2 light years, inches, etc. IE, 2 → 3 → 2 → 2 → 2 → 2 → 2 divided by 4 → 7 → 3 → 3. A six arrow number divided by a three arrow just returns the six, right? The best that can be done with chained arrow notation, I think. I gotta be correct about this, these aliens are supposed to be super-smart (: . 2→3→2→2→2→2 =2→3→2→2→(2→3→2→2→1→2)→1 =2→3→2→2→(2→3→2→2) =2→3→2→2→(2→3→(2→3→1→2)→1) =2→3→2→2→(2→3→(2→3)) =2→3→2→2→(2→3→8) =2→3→2→2→(2→(2→2→8)→7) =2→3→2→2→(2→4→7) =2→3→2→2→(2↑↑↑↑↑↑↑4) Rpakr (talk) 19:08, January 25, 2018 (UTC) You're missing an →2; its a six arrow number, heh. My point is, that 6-arrow 7 digit no. is so large, it can be divided by a number that dwarfs Graham's no, and the result is the same exact number, in chained arrow notation.Chasrob1950 (talk) 05:21, January 26, 2018 (UTC) Here's the number I'm talking about-- 2→3→2→2→2→2→2 =2→3→2→2→2→(2→3→2→2→2→1→2)→1 =2→3→2→2→2→(2→3→2→2→2) =2→3→2→2→2→(2→3→2→(2→3→2→1→2)→1) =2→3→2→2→2→(2→3→2→(2→3→2)) =2→3→2→2→2→(2→3→2→(2→(2→2→2)→1)) =2→3→2→2→2→(2→3→2→(2→4)) =2→3→2→2→2→(2→3→2→16) =2→3→2→2→2→(2→3→(2→3→1→16)→15) =2→3→2→2→2→(2→3→8→15) etc. Chasrob1950 (talk) 05:31, January 26, 2018 (UTC) Ok, I was told--if I distribute N points(domains) throughout a p-dimensional ball(the bubble) uniformly and select a point/domain, the distance between that point/domain and the nearest point/domain, on average, is R * (1 - 1/2^(1/N))^(1/p). N equals 4→7→3→3 in this case (R = 2→3→2→2→2→2→2). Ergo, 2 is raised to the (4→7→3→3)'th root, correct? 1.00000...vast no. of zeros...000(few more integers). Damned close to exactly one, for most all intents and purposes. So the result in brackets is more or less zero, and the equ. more or less equals zero. No help to me. ( Unless i consider R to be more or less infinite, so (2→3→2→2→2→2→2) times zero is 2→3→2→2→2→2→2. (: ) AIUI, 2→3→2→2→2→2→2 is the minimal 6 arrow, non-trivial chained arrow notation. Could i reasonably have the alien say the nearest domain is (googol) → (googolplex+3) → (Graham' Number +2) → (Friedman's n=4) → (googol +2) → (Skewes larger number +12)? IE, a very large 5 arrow number?Chasrob1950 (talk) 15:20, January 26, 2018 (UTC) Actually, no (if anyone cares (: ). The universe cannot be that anthropomorphic. Also, its colossi. Chasrob1950 (talk) 16:28, January 29, 2018 (UTC)